Skip to main content

computational geometry - Smooth convex hull of a large data set of 3D points


I recently needed to deal with a large data set of 600,000 points in three dimensions. My task was to find a convex hull for this data. I have Mathematica 10, so I could use the function ConvexHullMesh; I obtained this:


screenshot convex hull



I was wondering if there is some way to find a smooth convex hull (maybe an ellipsoid) for my data.



Answer



Minimum Volume Ellipsoid


Translated from here, this uses the Khachiyan algorithm, and should work for any dimension.


    MinVolEllipse[P_, tolerance_] := 
     Module[{d, n, Q, count, err, u, X, M, maximum, j, stepSize, newu, U, A, c},

      {n, d} = Dimensions[P];
      Q = Append[1] /@ P;


      count = 1;
      err = 1;
      u = ConstantArray[1./n, n];

      While[err > tolerance,
       X = Q\[Transpose].DiagonalMatrix[u].Q;
       M = Diagonal[Q.Inverse[X].Q\[Transpose]];
       maximum = Max[M];
       j = Position[M, maximum][[1, 1]];
       stepSize = (maximum - d - 1)/((d + 1) (maximum - 1));

       newu = (1 - stepSize) u;
       newu[[j]] += stepSize;
       count += 1;
       err = Norm[newu - u];
       u = newu;
       ];

      U = DiagonalMatrix[u];

      A = (1/d) Inverse[P\[Transpose].U.P - Outer[Times, u.P, u.P]];

      c = u.P;

      {c, A}
      ]

Usage:


    pts = RandomVariate[
       MultinormalDistribution[RandomReal[{-1, 1}, {2}], 
        With[{m = RandomReal[{0, 1}, {2, 2}]}, m.m\[Transpose]]], 500];


    P = MeshCoordinates[ConvexHullMesh[pts]];
    tolerance = 0.0001;

    {c, A} = MinVolEllipse[P, tolerance];

    X = {x, y};
    Show[
     ConvexHullMesh[pts],
     Graphics[{
       Point[pts],

       {Red, Point[P]}
       }],
     ContourPlot[(X - c).A.(X - c) == 1, {x, -4, 4}, {y, -4, 4}]
     ]

enter image description here


In 3D:


    pts = RandomVariate[
       MultinormalDistribution[RandomReal[{-1, 1}, {3}], 
        With[{m = RandomReal[{0, 1}, {3, 3}]}, m.m\[Transpose]]], 100];


    P = MeshCoordinates[ConvexHullMesh[pts]];
    tolerance = 0.0001;

    {c, A} = MinVolEllipse[P, tolerance];

    X = {x, y, z};
    Show[
     ConvexHullMesh[pts, MeshCellStyle -> {{2, All} -> Opacity[0.5, Green]}],
     Graphics3D[{

       Point[pts],
       {Red, Point[P]}
       }],
     ContourPlot3D[(X - c).A.(X - c) == 1, {x, -3, 3}, {y, -3, 3}, {z, -3, 3},
      ContourStyle -> Directive[Red, Opacity[0.2]]
      ]
     ]

enter image description here


Comments

Post a Comment

Popular posts from this blog

front end - keyboard shortcut to invoke Insert new matrix

I frequently need to type in some matrices, and the menu command Insert > Table/Matrix > New... allows matrices with lines drawn between columns and rows, which is very helpful. I would like to make a keyboard shortcut for it, but cannot find the relevant frontend token command (4209405) for it. Since the FullForm[] and InputForm[] of matrices with lines drawn between rows and columns is the same as those without lines, it's hard to do this via 3rd party system-wide text expanders (e.g. autohotkey or atext on mac). How does one assign a keyboard shortcut for the menu item Insert > Table/Matrix > New... , preferably using only mathematica? Thanks! Answer In the MenuSetup.tr (for linux located in the $InstallationDirectory/SystemFiles/FrontEnd/TextResources/X/ directory), I changed the line MenuItem["&New...", "CreateGridBoxDialog"] to read MenuItem["&New...", "CreateGridBoxDialog", MenuKey["m", Modifiers-...
Post a Comment
Read more

How to thread a list

I have data in format data = {{a1, a2}, {b1, b2}, {c1, c2}, {d1, d2}} Tableform: I want to thread it to : tdata = {{{a1, b1}, {a2, b2}}, {{a1, c1}, {a2, c2}}, {{a1, d1}, {a2, d2}}} Tableform: And I would like to do better then pseudofunction[n_] := Transpose[{data2[[1]], data2[[n]]}]; SetAttributes[pseudofunction, Listable]; Range[2, 4] // pseudofunction Here is my benchmark data, where data3 is normal sample of real data. data3 = Drop[ExcelWorkBook[[Column1 ;; Column4]], None, 1]; data2 = {a #, b #, c #, d #} & /@ Range[1, 10^5]; data = RandomReal[{0, 1}, {10^6, 4}]; Here is my benchmark code kptnw[list_] := Transpose[{Table[First@#, {Length@# - 1}], Rest@#}, {3, 1, 2}] &@list kptnw2[list_] := Transpose[{ConstantArray[First@#, Length@# - 1], Rest@#}, {3, 1, 2}] &@list OleksandrR[list_] := Flatten[Outer[List, List@First[list], Rest[list], 1], {{2}, {1, 4}}] paradox2[list_] := Partition[Riffle[list[[1]], #], 2] & /@ Drop[list, 1] RM[list_] := FoldList[Transpose[{First@li...
Post a Comment
Read more

plotting - Magnifying Glass on a Plot

Although there is a trick in TEX magnifying glass but I want to know is there any function to magnifying glass on a plot with Mathematica ? For example for a function as Sin[x] and at x=Pi/6 Below, this is just a picture desired from the cited site. the image got huge unfortunately I don't know how can I change the size of an image here! Answer Insetting a magnified part of the original Plot A) by adding a new Plot of the specified range xPos = Pi/6; range = 0.2; f = Sin; xyMinMax = {{xPos - range, xPos + range}, {f[xPos] - range*GoldenRatio^-1, f[xPos] + range*GoldenRatio^-1}}; Plot[f[x], {x, 0, 5}, Epilog -> {Transparent, EdgeForm[Thick], Rectangle[Sequence @@ Transpose[xyMinMax]], Inset[Plot[f[x], {x, xPos - range, xPos + range}, Frame -> True, Axes -> False, PlotRange -> xyMinMax, ImageSize -> 270], {4., 0.5}]}, ImageSize -> 700] B) by adding a new Plot within a Circle mf = RegionMember[Disk[{xPos, f[xPos]}, {range, range/GoldenRatio}]] Show...
Post a Comment
Read more